Question

A piston-cylinder device contains steam that undergoes a reversible thermodynamic cycle. Initially the steam is at \(400 \mathrm{kPa}\) and \(350^{\circ} \mathrm{C}\) with a volume of \(0.3 \mathrm{m}^{3} .\) The steam is first expanded isothermally to \(150 \mathrm{kPa}\), then compressed adiabatically to the initial pressure, and finally compressed at the constant pressure to the initial state. Determine the net work and heat transfer for the cycle after you calculate the work and heat interaction for each process.

Short answer

#Calculations# First, we calculate the mass of the steam: \[m = \frac{p_1 V_1}{R T_1} = \frac{400 \times 0.3}{0.4615 \times 623.15} = 0.822\,\mathrm{kg}\] Assuming the given values for specific heat capacities of steam, \(c_p = 1.996\,\frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\) and \(c_v = 1.534\,\frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\), we can calculate the specific heat ratio, \(\gamma\): \[\gamma = \frac{c_p}{c_v} = \frac{1.996}{1.534} = 1.301\] #Step 2: Isothermal Expansion# Since the process is an isothermal expansion, the temperature during the expansion remains constant at \(T_1 = 623.15\,\mathrm{K}\). The final pressure is given as \(p_2 = 200\,\text{kPa}\). To find the final volume, we can use the ideal gas equation: \[p_1 V_1 = p_2 V_2\] Thus, the final volume is: \[V_2 = \frac{p_1 V_1}{p_2} = \frac{400 \times 0.3}{200} = 0.6\,\mathrm{m^3}\] We can now calculate the work done during the isothermal expansion as: \[W_{1\to 2} = m R T_1 \ln\frac{V_2}{V_1} = 0.822 \times 0.4615 \times 623.15 \times \ln\frac{0.6}{0.3} = 81.17\,\mathrm{kJ}\] The heat transfer during the isothermal expansion is equal to the work done: \[Q_{1\to 2} = W_{1\to 2} = 81.17\,\mathrm{kJ}\] #Step 3: Adiabatic Compression# During the adiabatic compression, there is no heat transfer (\(Q_{2\to 3} = 0\)). The final pressure after compression is \(p_3 = 400\,\text{kPa}\). Using the isentropic relations for adiabatic processes, we can find the final volume: \[\frac{p_2 V_2^\gamma}{\gamma - 1} = \frac{p_3 V_3^\gamma}{\gamma - 1} \Rightarrow V_3 = \left(\frac{p_2}{p_3}\right)^{\frac{1}{\gamma}}V_2 = 0.496\,\mathrm{m^3}\] The work done during the adiabatic compression is: \[W_{2\to 3} = \frac{m R (T_3 - T_2)}{\gamma - 1} = -77.49\,\mathrm{kJ}\] #Step 4: Isobaric Compression# The final process is an isobaric compression, which takes the system back to its initial state. Therefore, \(p_3 = p_1\) and \(V_3 = V_1 = 0.3\,\mathrm{m^3}\), and the work done in this process is: \[W_{3\to 1} = p_3(V_1 - V_3) = 400\times(0.3-0.496) = -78.4\,\mathrm{kJ}\] The heat transfer in the isobaric compression can be found using the first law of thermodynamics: \[Q_{3\to 1} = W_{3\to 1} + m c_v (T_1 - T_3) = 40.07\,\mathrm{kJ}\] #Net Work and Heat Transfer# Finally, we can calculate the net work and heat transfer for the cycle: Net work: \[W_{net} = W_{1\to 2} + W_{2\to 3} + W_{3\to 1} = 81.17 - 77.49 - 78.4 = -74.72\,\mathrm{kJ}\] Net heat transfer: \[Q_{net} = Q_{1\to 2} + Q_{2\to 3} + Q_{3\to 1} = 81.17 + 0 + 40.07 = 121.24\,\mathrm{kJ}\] #Answer# The net work for the three-step reversible thermodynamic cycle is approximately \(-74.72\,\mathrm{kJ}\), and the net heat transfer for the cycle is approximately \(121.24\,\mathrm{kJ}\).

Step-by-step solution

Initial State#EndTag#

Given the initial pressure \(p_1 = 400 \,\text{kPa}\), initial temperature \(T_1 = 350^\circ \mathrm{C} = 623.15\,\mathrm{K}\), and the initial volume \(V_1 = 0.3\,\mathrm{m^3}\) of the steam. We assume steam behaves as an ideal gas. First, we need to determine the initial mass and specific heat ratio for the steam. We can use the ideal gas equation: \[p_1 V_1 = m R T_1 \] We are given the specific gas constant for steam \(R = 0.4615 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\). We can use this to find the mass of the steam: \[m = \frac{p_1 V_1}{R T_1}\] Likewise, we can find specific heat capacities at constant pressure and volume for the steam, \(c_p\) and \(c_v\), and use this information to determine the specific heat ratio, \(\gamma\): \[\gamma = \frac{c_p}{c_v}\]

Key concepts

Isothermal Expansion

Isothermal expansion is a process where a gas expands at a constant temperature, meaning that the thermal energy of the system remains unchanged throughout the process. This is a key concept in thermodynamics and it is derived from the first law of thermodynamics, which states that the internal energy change in a system is equal to the heat added to the system minus the work done by the system. For an ideal gas undergoing isothermal expansion, any heat added to the gas is used to do work, expanding against an external pressure.

In the context of the piston-cylinder device from our exercise, the steam expands isothermally from its initial state to a lower pressure. The temperature remains constant at 350°C, which corresponds to 623.15 K.

• To calculate the work done during isothermal expansion, one uses the formula:
  \(W = mRT_1\ln\left(\frac{V_2}{V_1}\right)\)
• Since the expansion is isothermal, the heat transferred (Q) can be calculated by: \(Q = W\)

It's critical to understand that during isothermal processes, the system equilibrium is maintained and the process progresses infinitely slow, called a quasi-static process, allowing the temperature to remain constant.

Adiabatic Compression

Adiabatic compression occurs when a gas is compressed without any heat transfer between the gas and its environment. It is an adiabatic process, which is characterized by the principle that the change in internal energy of the system is equal to the work done on the system. Since there is no heat exchange, all the work done on the gas goes into increasing its internal energy, and consequently its temperature.

For a reversible adiabatic process involving an ideal gas, the relationship between pressure, volume, and temperature follows Poisson's Law, which states that:
\(pV^\gamma = \text{constant}\) and \(TV^{\gamma-1} = \text{constant}\), where \(\gamma\) is the heat capacity ratio, a term that involves specific heat capacities at constant pressure and volume.

The work done during the adiabatic compression can be determined using the formula:
\(W = \frac{p_2 V_2 - p_1 V_1}{\gamma -1}\), where \(p_1\), \(V_1\) are the initial pressure and volume, and \(p_2\), \(V_2\) are the final pressure and volume after compression. In this process, since there's no heat exchange, \(Q = 0\).

A key point to remember is that when dealing with adiabatic compression, it requires the assumption that the compression happens quickly so that no heat is lost to the surroundings.

Ideal Gas Law

The ideal gas law is a fundamental equation that describes the behavior of an ideal gas. This law relates the pressure, volume, temperature, and amount of substance of a gas in a simple formula:
\(pV = nRT\), where \(p\) stands for pressure, \(V\) is volume, \(n\) is the number of moles of gas, \(R\) is the universal gas constant, and \(T\) represents temperature in Kelvin. For practical calculations, it's common to replace the term \(n\) with \(m/M\), where \(m\) is the mass of the gas and \(M\) is its molar mass, leading to an alternative form of the ideal gas law:
\(pV = \frac{m}{M}RT\).

The ideal gas law is used as the basis for many calculations in thermodynamics. It is particularly useful because it allows us to calculate properties of a gas that is undergoing changes in pressure, volume, and temperature, under the assumption that the gas behaves ideally. This law implies that for an ideal gas, the molecules do not interact with each other and occupy no volume themselves, though this is not entirely accurate for real gases.

• In the exercise mentioned, steam is assumed to behave as an ideal gas, which allows the use of the ideal gas law to calculate the mass of the steam using the formula:
  \(m = \frac{p_1 V_1}{R T_1}\)
• It also helps in establishing relationships used in the calculation of work and heat for isothermal and adiabatic processes.

Understanding the ideal gas law is crucial when solving thermodynamic problems because it serves as the starting point for describing the behavior of gases under different thermodynamic processes.